Circles and Parabolas

Slicing a cone

Imagine cutting through a double cone with a flat knife. Depending on how you tilt the knife, the cut edge is a circle, an ellipse, a parabola or a hyperbola. These curves are the conic sections. After a quick look at all four, we settle down with two of them: the circle, with its standard and general equations, and the parabola, with its focus, directrix, axis and latus rectum.

Conic sections

Here is the precise picture. Take a double cone with vertical axis and semi-vertical angle $\alpha$ , cut by a plane making angle $\beta$ with the axis:

$\beta = 90^\circ$ : a circle;
$\alpha < \beta < 90^\circ$ : an ellipse;
$\beta = \alpha$ : a parabola;
$0 \le \beta < \alpha$ : a hyperbola.

If the plane happens to pass through the vertex, the cut collapses into a point, a line or a pair of lines.

The circle

You know a circle already: it is the set of points at a fixed distance $r$ from a fixed centre $(h, k)$ :

$(x - h)^2 + (y - k)^2 = r^2.$

Open out the brackets and you get the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ , with centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$ (for a real circle you need $g^2 + f^2 > c$ ).

Example 1. With centre $(-3, 2)$ and radius $5$ we get $x^2 + y^2 + 6x - 4y - 12 = 0$ .

Example 2. $x^2 + y^2 - 8x + 10y - 12 = 0$ . Complete the squares: $(x - 4)^2 + (y + 5)^2 = 53$ , so the centre is $(4, -5)$ and the radius is $\sqrt{53}$ .

Example 3. The circle through $(0, 0)$ , $(6, 0)$ and $(0, 8)$ . Since the origin lies on it, $c = 0$ . Then $36 + 12g = 0$ gives $g = -3$ ; $64 + 16f = 0$ gives $f = -4$ : $x^2 + y^2 - 6x - 8y = 0$ , centre $(3, 4)$ , radius $5$ .

Circle with centre (3, 4) and radius 5 passing through the three points (0, 0), (6, 0) and (0, 8); one radius is drawn from the centre to (6, 0). — Example 3: the circle through (0, 0), (6, 0) and (0, 8)

Example 4. Suppose the centre lies on the line $x + y = 4$ , and the circle passes through $(2, 5)$ and $(6, 1)$ . Both points are the same distance from the centre, so $(h - 2)^2 + (k - 5)^2 = (h - 6)^2 + (k - 1)^2$ gives $h - k = 1$ ; with $h + k = 4$ : $\displaystyle h = \tfrac{5}{2}$ , $\displaystyle k = \tfrac{3}{2}$ ; $\displaystyle r^2 = \tfrac{1}{4} + \tfrac{49}{4} = \tfrac{25}{2}$ .

The parabola

A parabola is the set of points that stay equally far from a fixed point (the focus) and a fixed line (the directrix). With focus $(a, 0)$ and directrix $x = -a$ ( $a > 0$ ):

$y^2 = 4ax.$

Here the axis is the $x$ -axis and the vertex is the origin. The latus rectum is the chord through the focus perpendicular to the axis; its length is $4a$ . The table below shows how the picture changes as the parabola turns to face other directions.

Equation	Focus	Directrix	Opens
$y^2 = 4ax$	$(a, 0)$	$x = -a$	right
$y^2 = -4ax$	$(-a, 0)$	$x = a$	left
$x^2 = 4ay$	$(0, a)$	$y = -a$	up
$x^2 = -4ay$	$(0, -a)$	$y = a$	down

Example 5. $y^2 = 12x$ : $a = 3$ ; focus $(3, 0)$ , directrix $x = -3$ , latus rectum $12$ .

Parabola y squared = 12x opening to the right from the origin, with focus (3, 0), the dashed directrix x = -3, and the latus rectum from (3, -6) to (3, 6), of length 12. — Example 5: focus, directrix and latus rectum of y squared = 12x

Example 6. $x^2 = -10y$ : $\displaystyle a = \tfrac{5}{2}$ ; focus $\displaystyle (0, -\tfrac{5}{2})$ , directrix $\displaystyle y = \tfrac{5}{2}$ .

Example 7. Vertex $(0, 0)$ , axis the $y$ -axis, through $(4, -2)$ : $x^2 = -4ay$ with $16 = 8a$ : $x^2 = -8y$ .

Example 8. The cable of a suspension bridge hangs as a parabola, with $100$ m between towers, and its lowest point $10$ m below the tower tops. Put the vertex at the origin; then $x^2 = 4ay$ through $(50, 10)$ : $x^2 = 250y$ .

Bridge cable hanging as the parabola x squared = 250y between two towers 100 m apart; its lowest point is the vertex (0, 0), 10 m below the tower tops, and it meets a tower at (50, 10). — Example 8: the bridge cable with its vertex at the origin

Try these yourself

Find the circle with centre $(1, -2)$ and radius $4$ ; with centre $(-a, -b)$ and radius $\sqrt{a^2 - b^2}$ .
Find the centre and radius of each: $x^2 + y^2 + 4x - 6y = 3$ ; $2x^2 + 2y^2 - x = 0$ .
Find the circle through $(2, 3)$ and $(-1, 1)$ with centre on $x - 3y - 11 = 0$ .
Find the circle with radius $5$ , centre on the $x$ -axis, passing through $(2, 3)$ .
Does $(-2.5, 3.5)$ lie inside, on or outside $x^2 + y^2 = 25$ ?
Find the focus, directrix and latus rectum of each: $y^2 = 8x$ ; $x^2 = 6y$ ; $y^2 = -16x$ .
Find the parabola with focus $(0, -4)$ and directrix $y = 4$ .
Find the parabola with vertex $(0, 0)$ , axis along the $x$ -axis, through $(3, -6)$ .
Find the area of the triangle formed by the vertex of $x^2 = 12y$ and the ends of its latus rectum.
An arch is a parabola with vertical axis, $8$ m wide at the base and $6$ m high. How wide is it $2$ m below the top?

Answers to check against

Show answers

$x^2 + y^2 - 2x + 4y - 11 = 0$ ; $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$ .
$(-2, 3)$ , $4$ ; $\displaystyle (\tfrac{1}{4}, 0)$ , $\displaystyle \tfrac{1}{4}$ .
$(h - 2)^2 + (k - 3)^2 = (h + 1)^2 + (k - 1)^2$ gives $6h + 4k = 11$ ; with $h - 3k = 11$ : $\displaystyle h = \tfrac{7}{2}$ , $\displaystyle k = -\tfrac{5}{2}$ ; $\displaystyle r^2 = \tfrac{65}{2}$ : $x^2 + y^2 - 7x + 5y - 14 = 0$ .
$(h - 2)^2 + 9 = 25$ : $h = 6$ or $-2$ : $(x - 6)^2 + y^2 = 25$ or $(x + 2)^2 + y^2 = 25$ .
$6.25 + 12.25 = 18.5 < 25$ , so the point is inside.
$(2, 0)$ , $x = -2$ , $8$ ; $\displaystyle (0, \tfrac{3}{2})$ , $\displaystyle y = -\tfrac{3}{2}$ , $6$ ; $(-4, 0)$ , $x = 4$ , $16$ .
$x^2 = -16y$ .
$y^2 = 12x$ .
Ends $(\pm 6, 3)$ : area $\displaystyle \tfrac{1}{2} \cdot 12 \cdot 3 = 18$ .
Put the vertex at the top: $x^2 = -4ay$ through $(4, -6)$ : $16 = 24a$ ; at $y = -2$ : $\displaystyle x^2 = \tfrac{16}{3}$ , width $\displaystyle 2\sqrt{\tfrac{16}{3}} = \tfrac{8}{\sqrt{3}} \approx 4.62$ m.

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Slic­ing a cone#

Conic sec­tions#

The cir­cle#

The parabola#

Try these your­self#

Answers to check against#

Test yourself